The activation energy of an uncatalyzed reaction is 95 kJ/mol. The addition of a catalyst lowers the activation energy to 55 kJ/mol. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (b) 125 °C?

Video transcript

Hi everyone. This problem reads an ester verification reaction has an activation barrier of 19.5 kg per mole at 25°C. The activation barrier is lowered to 12.3 kg per mole when a catalyst is added to the reaction mixture. If the frequency factor for the reaction does not change, how many times faster is the catalyzed reaction than the un catalyzed reaction? So our goal here is to figure out how many times faster okay, is are catalyzed reaction than the un catalyzed reaction? To calculate the ratio by which the rate of reaction increases, we can calculate the ratio of the rate constants and the way that we're going to do that is by using this equation, Ln of K is equal to negative activation energy over R. T plus Ln of A. So let's take a look at what we have. So we'll say K one is equal to our uncapped realized K two is equal to our catalyzed reaction. Okay, now, our first activation energy Is equal to 19.5 kg joules per mole. Our 2nd activation energy Is equal to 12.3. Kill jules per mole. R is a constant. We should know and have memorized and that is 8. jewels over mole times Kelvin. So now if we take a look, okay, let's go ahead and write the temperature out too. So temperature is equal to 25°C. So this is all that we're given and we need to make sure our units match. So if we take a look at our our constant, we can see that our energy is in jewels and our temperature is in Kelvin. So that means we need to convert our activation energies and our temperature to jewels and Kelvin. So let's go ahead and do that. So in one kill a jewel There is 1000 jewels. So that makes our first activation energy Equal to 19, jewels per mole. And that makes our second activation energy equal to 12,300 jewels per mole. So now we need to convert our temperature to Kelvin and from to go from C to Kelvin, we add 273.15. So this gives us a temperature of 298.15 Kelvin. So now all of our units match and we're able to now solve. Okay, so the way that we're going to do this is we're going to write this equation out as Ln let's move this down a little bit. So we're going to have Ln of K 2 -6 of K one is equal to negative activation energy to over R. T plus Ln of a minus activation energy. One over R T plus Ellen of a. Okay, so we're going to calculate this so that we can find out our ratio of the rate constants and we'll be able to solve. Okay, so let's go ahead and plug in what we know. So we have Ln of K two minus Ln of K one is equal to negative 12,300 jewels over 8. jules over mall times, kelvin times are temperature 298.15 kelvin plus Ln of A. So that's our first one and this is going to be minus negative 19,500 jewels over 8.314 jules over more times, kelvin Times. Kelvin plus Ellen of a. Okay, so we can go ahead and simplify this out by pulling and factoring out. Okay, so we'll get Ln of K two minus Ln of K one is equal to negative 12,300 jewels per mole plus 19,500 jewels per mole. And this is all over our denominator. So 8.314 jewels over mole times, kelvin times our temperature. 298.15 kelvin. Okay, so now that we did that this is going to create a ratio and our kelvin's here are going to cancel. And so our ratio becomes Ln of K two over Ln of K one is equal to 7200 jewels per mole over 2479 jewels per mole. Okay, so when we simplify this out, we get Ellen of K two over Ellen of K one is equal to two point 904. So we need to eliminate the L N. On the left side. And the way that we do that is by raising the right side two E to the 2.904. When we do that, our l. n. Is going to cancel. Okay? So this will cancel out And we're going to get a ratio of K2 over K1 is equal to 18.02. So the rate of the reaction is increased by a factor of 18.02. And that is the answer to this problem. And that's the end of this problem. I hope this was helpful.